将字符向量中的日期替换为特定格式

我得到以下字符向量：

"On the evening of 2017-04-23, I was too tired"
"to complete my homework that was due on 24.04.2017."

我需要搜索它以查找所有出现的日期，并将其替换为“Monthname D, YYYY”格式。

我知道一般格式应该是 %B %d, %Y 并且我可能必须使用sub()功能，但我不太确定如何将两者结合在一起。

当我尝试类似的事情时

sub("[0-9]{2}.[0-9]{2}.[0-9]{4}","%B %d, %Y",x)

我刚刚得到以下结果

"On the evening of 2001-01-15, I was too tired to complete my homework that was due on %B %d, %Y."

有人可以帮我弄清楚如何将它们整合在一起吗？

在 stackoverflowers 的帮助下我的新代码如下：

streamlineDates(x)
{
#set pattern to dates in form of YYYY-MM-DD or DD.MM.YYYY
pattern <- "\\d{2,4}[.-]\\d{2}[.-]\\d{2,4}"

y <- c(x)

val <- unlist(regmatches(y, gregexpr(pattern, y)))

val1 <- as.Date(val,format=c("%Y-%m-%d","%d.%m.%Y"))
val2 <- format(val1,"%B %d, %Y")

y1 <- list()
for (i in 1:length(y)){
  y1[i] <- gsub(pattern,val2[i],y[i])
}
}

但是，当我仅输入：

x <- "to complete my homework that was due on 24.04.2017."

...它只返回 NA。我已将问题范围缩小到gsub哪里的替代品 value, “如果 NA，则结果中与匹配项对应的所有元素都将设置为 NA”。因此，由于仅输入最后一行时缺少第一个日期，因此仅返回 NA。

我怎样才能让它接受其中一个或两个日期？

第一种方法：

Using BASE R解决方案（不使用任何包）：

pattern <- "\\d{2,4}[.-]\\d{2}[.-]\\d{2,4}"
rep <- c("On the evening of 2017-04-23, I was too tired","to complete my homework that was due on 24.04.2017.")


val <- unlist(regmatches(rep, gregexpr(pattern, rep)))

val1 <- as.Date(val,format=c("%Y-%m-%d","%d.%m.%Y"))
val2 <- format(val1,"%B %d, %Y")
val2
rep1 <- list()
for (i in 1:length(rep)){
rep1[i] <- gsub(pattern,val2[i],rep[i])
}

Answer:

do.call("c",rep1)

> do.call("c",rep1)                                                   
[1] "On the evening of April 23, 2017, I was too tired"      
[2] "to complete my homework that was due on April 24, 2017."
> 

第二种方法：

使用库stringr

library(stringr)
rep <- c("On the evening of 2017-04-23, I was too tired","to complete my homework that was due on 24.04.2017.")
val <- str_extract(rep,"\\d{2,4}[.-]\\d{2}[.-]\\d{2,4}")
val1 <- as.Date(val,format=c("%Y-%m-%d","%d.%m.%Y"))
val2 <- format(val1,"%B %d, %Y")
rep1 <- str_replace_all(rep,"\\d{2,4}[.-]\\d{2}[.-]\\d{2,4}",val2)
rep1

Answer:

> rep1
[1] "On the evening of April 23, 2017, I was too tired"      
[2] "to complete my homework that was due on April 24, 2017."
> 

编辑OP稍微改变了问题之后，解决方案更通用，但是假设月份始终位于中间，并且分隔符仅限于破折号（-）和点（.）：

pattern <- "\\d{2,4}[.-]\\d{2}[.-]\\d{2,4}"
rep <- c("On the evening of 2017-04-23, I was too tired","to complete my homework that was due on 24.04.2017.")


val <- unlist(regmatches(rep, gregexpr(pattern, rep)))

year <- regmatches(val, gregexpr("\\d{4}", val))

month <- regmatches(val, gregexpr("(?<=[.-])\\d{1,2}(?=[.-])", val,perl=T))

date <- regmatches(val, gregexpr("(?<=[.-])\\d{2}$|^\\d{2}(?=[.-])", val,perl=T))
#Extracting year month and date , assuming month always falls in middle string

date1 <- paste0(year,"-",month,"-",date)
date1 <- as.Date(date1,"%Y-%m-%d")
val2 <- format(date1,"%B %d, %Y")

rep1 <- list()
for (i in 1:length(rep)){
  rep1[i] <- gsub(pattern,val2[i],rep[i])
}


do.call("c",rep1)